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Dương Chí Thắng
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Huỳnh Diệu Linh
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your heart your love is...
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Nguyễn Phan Thục Trinh
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Nguyễn Minh Quang
9 tháng 2 2021 lúc 20:39

Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)

\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)

\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)

Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)

Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)

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Nhok Song Tử
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Nguyễn Huy Tú
11 tháng 12 2020 lúc 13:11

a, \(\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right):\left(x+\frac{x}{x-1}\right)\)

\(=\left(\frac{x^3+1}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)+x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left[x^2-x+1-x^2+1\right]}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2}{x-1}\right)\)

\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x^2}=\frac{2-x}{x^2}\)

b, Ta có : A = 3 hay  \(\frac{2-x}{x^2}=3\)

\(3x^2=2-x\Leftrightarrow3x^2+x-2=0\)

\(\Leftrightarrow3x^2+3x-2x-2=0\Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\Leftrightarrow x=-1;\frac{2}{3}\)

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Ngô Chi Lan
11 tháng 12 2020 lúc 15:22

a,\(A=\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right)\div\left(x+\frac{x}{x-1}\right)\)

\(=\left(\frac{x^3+1}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x^2-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\div\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x\left(x-1\right)+x}{\left(x-1\right)}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1-x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x^2}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{x^2}{x-1}\)

\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{x^2}\)

\(=\frac{\left(x+1\right)\left(2-x\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)x^2}=\frac{2-x}{x^2}\)

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ĐÀO THỊ HUYỀN DIỆU
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Duy đây rồi
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Nguyễn Hồng Thắm
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Nguyễn Hồng Thắm
6 tháng 10 2018 lúc 11:12

Ai giải giúp mình bài 1 với bài 4 trước đi

Quyết Tâm Chiến Thắng
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KAl(SO4)2·12H2O
1 tháng 2 2019 lúc 12:53

làm a thôi nha :D 

a) \(C=\left(\frac{x^2+x}{x^2-2x+1}\right):\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)

\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{2-x^2}{x\left(x+1\right)}\right]\)

\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right]\)

\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{x+2-x^2}{x\left(x-1\right)}\right]\)

\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{\left(x-1\right)\left(x+1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)

\(C=\frac{x+1}{x^2-2x+1}.\frac{x^2-1+x+2-x^2}{x-1}\)

\(C=\frac{x+1}{\left(x^2-2x+1\right)}.\frac{1.x}{x-1}\)

\(C=\frac{\left(x+1\right)^2}{x^3-x^2-2x^2+2x+x-1}\)

\(C=\frac{x^2+2x+1}{x^3-3x^2+3x-1}\)

Nguyệt
1 tháng 2 2019 lúc 13:20

a)\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)

\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1}{x.\left(x-1\right)}+\frac{x}{x.\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)

\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1+x-x^2+2}{x.\left(x-1\right)}\right]\)

\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x.\left(x-1\right)}\right]=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right].\left[\frac{x.\left(x-1\right)}{x+1}\right]=\frac{x.\left(x+1\right).x}{\left(x-1\right).\left(x+1\right)}=\frac{x^2}{x-1}\)

b)\(\text{Để B nguyên }\Rightarrow x^2⋮x-1\)

\(x^2=x^2-1+1=\left(x-1\right).\left(x+1\right)+1\)

\(\Rightarrow\text{Để }x^2⋮x-1\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\Rightarrow x\in\left\{2;0\right\}\)

nguyenthiluyen
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Trịnh Thành Công
27 tháng 6 2017 lúc 21:45

a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)

\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{x+1}{x}\right)+\frac{1}{\left(x+1\right)^2}.\left(\frac{x^2+1}{x^2}\right)\right]:\frac{x-1}{x^3}\)

\(P=\left[\frac{2}{\left(x+1\right)^2x}+\frac{x^2+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{x^3}\)

\(P=\left[\frac{x^2+2x+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{3}\)

\(P=\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}:\frac{x-1}{3}\)

\(P=\frac{3}{x^2\left(x-1\right)}\)

b)Bài này liên quan đến dấu lớn nên mk ko làm đc